Question

Use the sum or difference formula to determine the value of the trigonometric function. sin (-pi/12)

Answer 1

hint: pi/4 - pi/3 = -pi/12

Answer 2

thnx @jim_thompson5910

Answer 3

you're welcome

Answer 4

@jim_thompson5910, can you help me solve this? I got \[\sqrt{2}+\sqrt{6}\div2 \]

Answer 5

but i got it wrong

Answer 6

I got minus not addition

Answer 7

you should have

sin(-pi/12) = sin(pi/4 - pi/3)

sin(-pi/12) = sin(pi/4)cos(pi/3) - cos(pi/4)sin(pi/3)

do you have that as part of your steps?

Answer 8

yass

Answer 9

so this is the full step by step picture you should have (or something similar to it)

sin(-pi/12) = sin(pi/4 - pi/3)

sin(-pi/12) = sin(pi/4)cos(pi/3) - cos(pi/4)sin(pi/3)

sin(-pi/12) = (sqrt(2)/2)*(1/2) - (sqrt(2)/2)*(sqrt(3)/2)

sin(-pi/12) = sqrt(2)/4 - sqrt(6)/4

sin(-pi/12) = (sqrt(2) - sqrt(6))/4

Therefore, \[\Large \sin\left(-\frac{\pi}{12}\right) = \frac{\sqrt{2}-\sqrt{6}}{4}\]

Answer 10

On step 3, I got

sin(pi/4) = sqrt(2)/2
sin(pi/3) = sqrt(3)/2
cos(pi/4) = sqrt(2)/2
cos(pi/3) = 1/2

using the unit circle

Answer 11

omg, i forgot its 4,i got all the work but i didn't multiply the 2. thnx

Answer 12

I gotcha, yw